The statement is true for all sentence symbols, as for a sentence symbol $A$ , either $A$ or $\neg A$ is in $\Delta$ , and $\overline{v}(A)=v(A)=T$ iff $A\in\Delta$ . Now, suppose it is true for any two wffs $\alpha$ and $\beta$ (that is, the evaluation of each one is $T$ or $F$ , depending on whether it or its negation is in $\Delta$ , respectively). Then, $\overline{v}(\alpha)=T$ iff $\alpha\in\Delta$ , hence, $\overline{v}(\neg\alpha)=T$ iff $\overline{v}(\alpha)=F$ iff $\alpha\notin\Delta$ , and, according to (ii), $\overline{v}(\neg\alpha)=T$ iff $(\neg\alpha)\in\Delta$ . Now, for $\alpha\square\beta$ , let $\neg?\alpha$ be $\alpha$ if $\alpha\in\Delta$ (iff $\overline{v}(\alpha)=T$ ) and $\neg\alpha$ otherwise (iff $\overline{v}(\alpha)=F$ ). Note, that $\neg\alpha\in\Delta$ iff $\alpha\notin\Delta$ , i.e. $\neg?\alpha\in\Delta$ and $\overline{v}(\neg?\alpha)=T$ . Similarly we define $\neg?\beta\in\Delta$ . Then, $\{\neg?\alpha,\neg?\beta,\neg?(\alpha\square\beta)\}$ , where $\neg?(\alpha\square\beta)$ means either $(\alpha\square\beta)$ or its negation depending on which one is in $\Delta$ (according to (ii), one of them is in $\Delta$ ), is a finite subset of $\Delta$ , therefore, by (i), it must be satisfiable. Now, for every truth assignment $w$ , $\overline{w}(\neg?(\alpha\square\beta))$ is uniquely determined by $\overline{w}(\alpha)$ and $\overline{w}(\beta)$ which are uniquely determined by $\overline{w}(\neg?\alpha)$ and $\overline{w}(\neg?\beta)$ . In particular, every truth assignment that satisfies $\{\neg?\alpha,\neg?\beta\}$ must satisfy $\neg?(\alpha\square\beta)$ , because if one such truth assignment satisfy it, so do all others (again, $\overline{w}(\neg?(\alpha\square\beta))$ is uniquely determined by $\overline{w}(\neg?\alpha)$ and $\overline{w}(\neg?\beta)$ ). Since $v$ satisfies both $\neg?\alpha$ and $\neg?\beta$ , it must satisfy $\neg?(\alpha\square\beta)$ , i.e. $\overline{v}(\alpha\square\beta)=T$ if $(\alpha\square\beta)\in\Delta$ , or $\overline{v}(\neg(\alpha\square\beta))=T$ if $\neg(\alpha\square\beta)\in\Delta$ , in which case $\overline{v}(\alpha\square\beta)=F$ . So, $\overline{v}(\alpha\square\beta)=T$ iff $(\alpha\square\beta)\in\Delta$ . By the induction principle, we conclude that for every wff $\phi$ , $\overline{v}(\phi)=T$ iff $\phi\in\Delta$ .